Start: Lecture #8, 11/5/2009In Si, about 2.25x1020at room temperatureIn Si, about 15 1010/31.5x1010e/cm3at room temperatureHow to determine thedetermine the Fermi level?Derivation of the Einstein Relationship:Relationship:This relationship can be used to determine the mobility from the ydiffusion coefficient and vice versa.Drift and Diffusion of carriersDrift DiffusionIn an electric field, hfFor electronshlthe current of charged carriers is controlled by both For holesydrift and diffusion. The sum of these two componentstwo components will determine how much current will flflowWhen we put p- and n-doped semiconductors in contact, the Fermi level of bothp-doped n-dopedpnpnthe Fermi level of both semiconductors has to become equal. Individual pieces pnpnThis results in the bending of of semiconductorgthe conduction and valence bands (band bending) and the establishment of an internalestablishment of an internal electrostatic potential VoWhen no current is flowing throught the junction, J(drift)+J(diffusion) = 0At equilibrium, when no current is flowing:Rearranging terms, we get:Substituting /D = q/kT (Einstein relationship):And finally, if we integrate, we get:This is the built-in potential VoppSince np=ni2, pnExample: A Si p-n junction has Na=1017on the p side and Nd=1016on the n-side. At 300K, what are the FiLldfidVFermi Levels and find Vo:Eip–Ef= kT ln pp/niipfppi= 0.0259 ln (1017/1.5x1010)=0.407 eV+ + + + + + -----xp00 xn0P-dopedN-dopedxTh h t b l b+ + + + + +--------N-dopedThere have to be an equal number of charges on either side of the junctionWN h hi l k lNote that this looks more or less like a capacitor with opposite charges on either side of an insulating regionggCharges must equal on both sides of junctionIf it’s a capacitor, the field depends on the dielectric constant and the number of charges (Poisson’s equation)Simplifying a bit, we get:we get:Next, we integrate to get the field:and W=xp0+xn0sincexn0=WNa/(Na+Nd), we can substitute:VoThis provides us with an equation which we can use to calculate the depletion width if we know the pdopant concentrations of a p-n junctionDepletion width equation:Depletion width Built-in potential permittivity of semiconductorpppywhere Naand Ndare the Acceptor and Donor Dopant concentrationsNote: For silicon, , the permittivity, is i(14/)Donor Dopant concentrationsgiven as 11.8 x (8.85x10-14F/cm)Dielectric constantPermittivity of free spaceconstantfree spaceWe can also determine how much of the depletion width is on the p and n side:Example: A Si diode is formed from an abrupt junction with Depletion widthNd=1016and Na=4x1018. Calculate Vo, xn0, xp0, Q+and the field for this junctionBuiltin potential VBuilt-in potential VoThis relationship is for no external voltage applied to the diode. If we apply a reverse bias voltage Va, we add that voltage to Voand use (Va+Vo) instead of Vohere. Forward bias reduces Wreduces WDepletion width Built-in potential permittivity of semiconductorwhere Naand Ndare the Acceptor and Donor Dopant concentrationsNote: For silicon, , the permittivity, is given as 11.8 x (8.85x10-14F/cm)Note: As we increase the reverse bias voltageNote: As we increase the reverse bias voltage, the depletion width increases.At positive voltages, the current follows an exponential pnpdependency on voltageThe Boltzman diode equationAt 0 li d V lt th tAt 0 applied Voltage, the current flowing is 0At negative voltages, the current saturates to a saturation current IgenSchematic of the junction terminology Position of the metallurgical (diffused) junctionpnExcess holes in Excess electrons usedthe n-materialin the p-materialDepletion widthHow much current flows through a diode when we apply a voltageWe know that the concentrations of holes in the p side and the concentration of holes on the n side of a p-n junction pjare related by Vo:Note that this means that as you apply a positive voltage the number of excess holespositive voltage, the number of excess holes on the n side increases exponentiallyNow, we can find the excess minority carrier concentration on both p and n-side of the depletion width by subtracting:width by subtracting:Since we know that the excess carrier concentration drops off to zero as we get further away from the junction, we can define diffusion lengths Lnand Lpfor l t d h l ti l d ti l d ith di telectrons and holes, respectively and assume an exponential decay with distance. These diffusion lengths depend on the dopant concentration (recombination).Usually the n and p regions are long with respect to Land Land:Usually, the n and p regions are long with respect to Lnand Lpand:Diffusion profile of excess minority carriersThese are the excess hole and electron concentrations as a function of distance xnand xpfrom the junctionWe remember the current density equation, and can simplify it to only contain the diffusion currentFrom this we can define theA is the crossectionalFrom this, we can define the diffusion current as:A is the crossectional area of the junctionThe total hole current is:The total hole current is:The total electron current isFinally, the total diode current is the sum of the hole and electron currents across the p-n junction and is given by:and is given by:Area of junction diffusivity diffusion length Applied voltage Temperature (K)L=DL=DCan also be substituted for LFor those interested, here are three ways of deriving the diodeof deriving the diode equationEnd: Lecture #8,